CAN YALL GO FOLLOW MY INSTAGRAM PLSS
2 answers:
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Answer:
(b)False
Explanation:
Given:
Prandtl number(Pr) =1000.
We know that
Where is the molecular diffusivity of momentum
is the molecular diffusivity of heat.
Prandtl number(Pr) can also be defined as
Where is the hydrodynamic boundary layer thickness and
is the thermal boundary layer thickness.
So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
So hydrodynamic layer will be thicker than the thermal boundary layer.
Answer:
a) the heat exchanger area required for the evaporator is 11178.236 m²
b) the required flow rate is 1993630.38 kg/s
Explanation:
Given the data in the question;
Water temperature near the surface = 300 K
temperature at reasonable depths ( cold ) = 280 K
power plant output W' = 2 MW
efficiency η = 3% = 0.03
we know that; efficiency η = W' / Q
we substitute
0.03 = 2 / Q
Q = 2 / 0.03
Q = 66.667 MW = 66.667 × 10⁶ Watt
T = 300 K T
= 292 K
T = 290 K T
= 290 K
Now, Heat transfer in evaporator;
Q = UA( LMTD )
so
LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )
first we get ΔT₁ and ΔT₂
ΔT₁ = T - T
= 300 - 290 = 10 K
ΔT₂ = T - T
= 292 - 290 = 2 K
so we substitute into our equation;
LMTD = (10 - 2) / ln( 10 / 2 )
LMTD = 8 / ln( 5 )
LMTD = 8 / 1.6094379
LMTD = 4.97
a) Heat transfer Area will be;
Q = UA( LMTD )
we substitute
66.667 × 10⁶ = 1200 × A × 4.97
66.667 × 10⁶ = 5964 × A
A = (66.667 × 10⁶) / 5964
A = 11178.236 m²
Therefore, the heat exchanger area required for the evaporator is 11178.236 m²
b) Flow rate
we know that;
Q = m'C
(
-
)
specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)
we substitute
66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )
66.667 × 10⁶ = m' × 33.44
m' = ( 66.667 × 10⁶ ) / 33.44
m' = 1993630.38 kg/s
Therefore, the required flow rate is 1993630.38 kg/s